I have an error in integrating to calculate the age of the Universe

In summary, there is an error in the process of integrating to calculate the age of the Universe. This error could be caused by inaccurate data or a flaw in the mathematical equations used for the calculation. Further investigation and refinement of the methods are needed in order to accurately determine the age of the Universe.
  • #36
Buzz Bloom said:
I do not understand why the integrand I uses is not equivalent to the integrand you posted.
Just look at them. But first you have to get them right.

Buzz Bloom said:
Mine is
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
Yes, that's what you posted earlier.

Buzz Bloom said:
Yours is based on
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
No, it isn't. I think you mistakenly quoted yourself again. My integrand is this:

PeterDonis said:
$$dt = \frac{1}{H_0} \frac{da}{\sqrt{\frac{M}{a} + L a^2}}$$
These are obviously not the same function. The integration variable being ##x## vs. ##a## is a cosmetic detail, the name of the variable doesn't matter, what matters is the functional form of the integrand. Since in both cases, the range of integration is ##0## to ##1##, and the functions being integrated are different, obviously the numerical answers will be different. And, as you have already found, your integrand gives the wrong answer. Mine gives the right answer.

Evidently you made some mistake in the substitutions you made to get your integrand in ##x## from the original one. That's why you're getting the wrong answer.
 
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  • #37
Hi @PeterDonis:

I did make an error in Post #35. Is is OK with you if I fix it with edits until I decide I have it right?

I think for any post I write I need to preface it with
"THIS IS A WORK IN PROGRESS,"
so if I make an error I can correct it before someone has responded to my error before I fixed it.

Regards,
Buzz
 
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  • #38
Buzz Bloom said:
I did make an error in Post #35. Is is OK with you if I fix it with edits until decide I have it right?
No. As has already been mentioned, once a post has responses, you should not do this. It is very confusing to other participants in the thread. Just make a follow-up post quoting what was in error from the previous one, and correcting it.

Buzz Bloom said:
I think for any post I write I need to preface it with
"THIS IS A WORK IN PROGRESS,"
so if I make an error I can correct it before someone has responded to my error before I fixed it.
This basically means nobody can ever respond to your posts until we're sure you can't edit them any more. That's not going to be feasible for other thread participants.

There is a "preview" function in the post window; I suggest that you use it again and again to review what you are going to post before you post it. The site will save your draft post so there is no need to post it just to have it saved. You can keep a post in draft for a while while you edit, preview, edit, preview, etc. until you are confident it is as good as you can get it, then post it.
 
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  • #39
Buzz Bloom said:
I did make an error in Post #35. Is is OK with you if I fix it with edits until decide I have it right?
Rather than go round and round with this, why don't you just respond to what I said in post #36?
 
  • #40
Hi @PeterDonis:

Well, I finally completed #19, and now have shown the process that produces a correct numerical result for the integral.

Regarding #36, your equation

$$dt = \frac {1}{H_0} {\frac {da} { \sqrt {M/a+L a^2}}} $$

becomes

$$dt = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$

by multiplying your numerator and denominator by ##1/{\sqrt {a}}##.

Then I substitute ##a^{1/3} = x## and take L outside of the sqrt. Both x and a have the integrations limits of 0 and 1. I also define the constant Q = M/2L.

My error (which I fixed) was not in the form of the equation, but in my calculations of the numerical arithmetic.

Thank you for helping me get used to LaTeX.

Regards,
Buzz
 
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  • #41
Hi @George Jones:

Re #33, my intent was to use the equation approach I presented (which you quoted in #33) to obtain a numerical result of about 13.8 giga-years. I finally was successful in Post #19. I was not trying to find a different way to integrate. Thank you for your interest.

Regards,
Buzz
 

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